Residue Theorem integral application


by jsi
Tags: complex analysis, integration, residue theorem
jsi
jsi is offline
#1
Apr3-12, 10:45 PM
P: 24
1. The problem statement, all variables and given/known data

Compute the integral: ∫ x2/(x4-4x2+5)

2. Relevant equations

Uses Residue theorem.

3. The attempt at a solution

So I found the zeroes of x4-4x2+5 to be 2+i and 2-i, and therefore the one that is of relevance is 2+i since it is in the upper half-plane. Then I used residue theorem that said Res(P(z)/Q(z); 2+i) = P(2+i)/Q'(2+i) = (2+i)2/(4(2+i)3-8(2+i)) = (4i+3)/(36i-8) and then I multiplied by 2∏i which would leave me with a value in the complex plane. I think this is wrong because it should come out with a real valued number. Does it have something to do with the zero having a multiplicity of 2? And if so, how do I go about redoing it with that in mind, I don't remember learning how to do that...
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clamtrox
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#2
Apr4-12, 05:57 AM
P: 937
Quote Quote by jsi View Post
So I found the zeroes of x4-4x2+5 to be 2+i and 2-i, and therefore the one that is of relevance is 2+i since it is in the upper half-plane.
Fourth order equation has four zeroes :)
jackmell
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#3
Apr4-12, 09:00 AM
P: 1,666
Quote Quote by jsi View Post
1. The problem statement, all variables and given/known data

Compute the integral: ∫ x2/(x4-4x2+5)

2. Relevant equations

Uses Residue theorem.

3. The attempt at a solution

So I found the zeroes of x4-4x2+5 to be 2+i and 2-i, and therefore the one that is of relevance is 2+i since it is in the upper half-plane. Then I used residue theorem that said Res(P(z)/Q(z); 2+i) = P(2+i)/Q'(2+i) = (2+i)2/(4(2+i)3-8(2+i)) = (4i+3)/(36i-8) and then I multiplied by 2∏i which would leave me with a value in the complex plane. I think this is wrong because it should come out with a real valued number. Does it have something to do with the zero having a multiplicity of 2? And if so, how do I go about redoing it with that in mind, I don't remember learning how to do that...
Until you analyze the entire problem meticulously with a fine-tooth comb, it's not going to happen even when you get all four. First, I assume you want the indefinite integral:

[tex]\int_{-\infty}^{\infty} \frac{x^2}{x^4-4x^2+5}dx[/tex]

via the Residue Theorem. Then we could write:

[tex]\mathop\oint\limits_{C} \frac{z^2}{z^4-4z^2+5}dz=\int_{-\infty}^{\infty} \frac{x^2}{x^4-4x^2+5}dx+\lim_{R\to\infty} \int\limits_{\gamma}\frac{z^2}{z^4-4z^2+5}dz=2\pi i \sum \text{Res}f(z)[/tex]

Now, you understand all that? Every little bit of it? Without me having to explain what all the notation is right? When you do, then analyze every part of it in detail even that "excipient" leg of the contour that I assume goes to zero but don't know for sure cus' I haven't analyzed it meticulously, then determine which of the four roots are in the contour, compute the residue of the function there, then do the final sum. Bingo-bango and we're done.

Dick
Dick is offline
#4
Apr4-12, 10:08 AM
Sci Advisor
HW Helper
Thanks
P: 25,175

Residue Theorem integral application


2+i isn't even a root of your polynomial. sqrt(2+i) (for example) is.


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